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Chapter Fourteen Review Answers

1. The cap and stalk died when the foot was removed. When the cap was cut off, the foot and stalk were not affected, and the cap regenerated. The A. mediterranea foot still grew an A. mediterranea cap, even with an A. crenulata stalk grafted onto it. Results suggested hereditary information resided in the foot (where the nucleus was).

2. They transplanted a nucleus into an enucleated fertilized egg with the resulting growth of a frog. An enucleated fertilized egg without a nucleus transplanted into it was not viable. Steward took macerated carrot tissue and managed to regenerate carrot tissue from single carrot cells.

3. The survivors were the ones injected with nonvirulent, uncoated bacteria or heat-killed, polysaccharide coated bacteria. The dead mice were injected with virulent bacteria containing a polysaccharide coat, or with a combination of live, non-polysaccharide-coated bacteria and dead, polysaccharide-coated bacteria. His conclusion was that the information specifying the coat from heat-killed bacteria was transferred to the live, non-polysaccharide-coated bacteria.

4. The primary component was DNA as evidenced by the fact that it was digested by DNAse.

5. Hershey and Chase labeled viral nucleic acid and viral protein coat with separate radioactive isotopes. After infected cells were analyzed, they found only viral nucleic acid inside the infected cells, not viral protein coat. Fraenkel-Conrat performed essentially the same experiment, creating hybrid plant viruses with nucleic acid from one species and viral protein coat from another. Plants developed lesions from these hybrid viruses, but only of the kind specified by the nucleic acid, not the viral protein coat.

6. A DNA molecule is composed of repeating units of deoxyribose, which is a five carbon sugar; a phosphate group; and a nitrogenous base. The three-dimensional shape of the molecule is a double helical strand. Hydrogen bonding between nitrogenous bases is responsible for holding the two strands of the helix together. Purines can only hydrogen-bond with pyrimidines, which is what accounts for equal numbers of purines and pyrimidines in a DNA molecule.

7. One strand of a DNA molecule is a "mirror" of the other. Meselson and Stahl showed that DNA replication was semiconservative by labeling new DNA with a heavy isotope so that older and newer strands could be identified. After allowing the DNA to replicate a while, it was spun on a cesium gradient, showing the newly-replicated material to weigh intermediate between the labeled and unlabeled DNA, suggesting that one chain (the new one) was "heavy," while the older (original) chain was "light."

8. The leading strand is replicated continuously while the lagging strand is replicated in fragments (Okazaki fragments). The reason for the discrepancy in replication strategy for the different strands is that the two strands are anti-parallel.

9. In bacteria, the duplex is nicked at one site, and a strand on one or both sides is displaced, creating one or two replication forks. These proceed around the circle until complete. Eukaryotic DNA has numerous replication forks along each chromosome, working in discrete units of 10,000 to 1,000,000 base pairs in length.

10. Histones are puck-shaped proteins around which the DNA duplex winds to form bead-shaped wads called nucleosomes. Nucleosomes are further condensed into visible dark areas in cell nuclei called nuclei. Chromatin is DNA: heterochromatin is usually densely packed and therefore not transcribed. Euchromatin, on the other hand, is DNA in a diffuse enough form to have its genes transcribed during interphase.

11. Beadle and Tatum wanted to test that hypothesis that DNA coded for specific enzymes, and tested their hypothesis on mold. They irradiated mold to see if they could create mutants incapable of synthesizing necessary enzymes in certain biosynthetic pathways. They were successful. Some of their mutants could not grow on a minimal medium unless they were supplied with the missing enzyme.



 

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